Here, note that our nucleophile (the conjugate base of an alkyne, pKa 25) can remove the proton of an alcohol (pKa ~15) or perform an SN2 reaction on the primary alkyl halide. With a difference of 10 pKa units between the alkyne and the alcohol, the acid-base reaction between the deprotonated alkyne (“acetylide”, stronger base) to produce a deprotonated alcohol (“alkoxide”, weaker base) is extremely favorable. And since acid-base reactions are fast, relative to other reactions, the preferred first reaction here is deprotonation of the alcohol to give the conjugate base (“alkoxide”)

Bonus question: what would be the final product of this reaction, after the deprotonation? Answer below.

Example 2. Acid-base reaction versus addition to a carboxylic acid

Grignard reagents are very good nucleophiles – reacting with carbonyl compounds such as ketones, aldehydes, and esters. But as the conjugate bases of alkanes (pKa ~ 50) they are also extremely strong bases. When combined with a carboxylic acid (pKa ~4 or 5) the result is not an addition to the carbonyl, but an acid base reaction (45 pKa units makes for a pretty favorable reaction!). It’s always helpful to remember that carboxylic acids… are acids.

Example 3 – Another Grignard

The same applies for reactions of Grignard reagents with molecules that have hydroxyl groups in addition to aldehydes or ketones. If merely one equivalent is added, the first thing to happen will be deprotonation of the alcohol, which is faster than addition to the ketone carbonyl carbon. It’s only after addition of a second equivalent of Grignard that addition to the ketone will occur.

Acid-base reactions on “heteroatoms” (that means atoms other than carbon, such as O, N, and S) generally require very little reorganization of the nuclei in the structure. Therefore these reactions are fast, relative to reactions where the nuclei have to move or be reorganized.

Think about removing a proton from an O-H. After loss of hydrogen, the oxygen gains a new lone pair. But its hybridization doesn’t change – it started as sp3, and it’s still sp3. So the nuclei (other than the H, of course) don’t significantly change positions in these reactions. No extra motion, in other words.

However when bonds are formed or broken at carbon – such as in the SN2 reaction or in additions to carbonyl carbons – a lot of atomic furniture has to get rearranged. For instance, the SN2 proceeds through a backside attack, which means that the geometry of the molecule changes from tetrahedral to trigonal bipyramidal (that’s the 5-coordinate transition state) and then back to tetrahedral. In addition to carbonyl compounds, we’re changing the hybridization of carbon from sp2 to sp3. That requires a shift from trigonal planar to tetrahedral geometry.

P.S. It’s interesting that Grignard reagents (the conjugate bases of alkanes, pKa ~50) don’t usually deprotonate aldehydes (pKa ~18) or ketones (pKa ~20). That’s another application of this principle. Removing a proton from an aldehyde or ketone requires breaking a C-H bond, and the resulting base (called an “enolate”) will undergo a change in hybridization from sp3 to sp2. Hence, it’s slow.

P.P.S. After the first acid-base reaction, the deprotonated alcohol can then do an SN2 reaction on the primary alkyl bromide.

I have a question that’s been bothering me for a while now–mostly because my organic chem prof doesn’t explain concepts fully–when is the best time an alkyl or hydride shift can be done on a reaction mechanism problem? I know alkyl / halide shifts can be used in a problem where you have an alcohol and you’re reacting it with some type of haloalkane like HBr, but are there any other times the shifts can be used?

Forgot to add that yes, I realize the shifts can be used in SN1 / SN2 reactions…maybe it’s best if I keep an eye out for any potential carbocation rearrangements that could make a carbon center more stable.

Thank you so much for all your work on the website. I got orgo 1 summary sheet and finding it very helpful.
Just a question. Why breakage of C-H bond results in rehybridazation om aldehyde and ketone?
Wouldn’t it be still sp3 with the lone pair after deprotonation?
There is no
Thank you again.

Hi, if you deprotonate C-H adjacent to a ketone, the carbon will not be sp3 hybridized for long. It is energetically favourable for the carbon to re-hybridize to sp2, which puts the lone pair in a p orbital and can then be conjugated with the p orbitals in the C=O group.

I purchased the summary, and realized I wasn’t ready for summary yet, and going through the posts now, and it’s really helpful. I just thought it will be really good if you can put together these post contents into printable pdf. I would definitely pay a reasonable price to have this contents easier to read, carry, and make notes..!

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